Solving ODE rigorously I am given the ODE 
$$(f''(r)+\frac{f'(r)}{r})(1+f'(r)^2)-f'(r)^2f''(r)=0$$
and want to solve it rigorously for $r>0.$ So especially, I don't want to loose any solutions. 
$\textbf{Derivation of the solutions}$
First I multiply by $r$
$$(rf''(r)+f'(r))(1+f'(r)^2) =rf'(r)^2f''(r)$$
which is nothing but
$$\frac{d}{dr}(rf'(r))(1+f'(r)^2) =rf'(r)^2f''(r)$$
now assuming $f'(r) \neq 0$ (Problem 1) I can rewrite this as
$$\frac{1}{r f'(r)} \frac{d}{dr}(rf'(r)) = \frac{f'(r)f''(r)}{1+f'(r)^2}$$
Assuming $f'(r)>0$ (Problem 2) we get
$$\frac{d}{dr}\ln(rf'(r)) = \frac{1}{2} \frac{d}{dr} \ln(1+f'(r)^2)$$
which gives after integrating
$$\ln(r^2f'(r)^2) = ln(1+f'(r)^2)+C$$ for some constant $C.$
Exponentiating this gives
$$r^2 f'(r)^2 = \underbrace{e^C}_{=:C_1^2>0} (1+f'(r)^2).$$
or alternatively 
$$f'(r)^2(r^2-C_1^2) = C_1^2$$ 
Problem (3) $r \neq C_1$ and Problem (4) (taking $\pm$) 
$$f'(r) = \frac{ \pm C_1 }{\sqrt{r^2-C_1^2}} = \frac{ \pm 1 }{\sqrt{\frac{r^2}{C_1^2}-1}}$$
Integrating gives then $f(r) = \pm C_1 \cosh^{-1} (\frac{r}{C_1})+d$ where $d$ is arbitrary and $C_1 >0.$
$\textbf{Regarding the problems:}$ Now, I want to understand whether I can get rid of the 4 problems (small inaccuracies in the way I am solving this ODE.)
Regarding problem 1: The question is: Do I really have to assume $f'(r) \neq 0$ or can I avoid this somehow?
Regarding problem 2: If I assume $f'(r)<0$ somwhere, it is quite obvious that we can rewrite the left hand side as $$\frac{1}{-rf'(r)} \frac{d}{dr}(-rf'(r)) = \ln(-rf'(r))$$ which would be well-defined now. Now, since we eventually square the argument of the $\ln $anyway (two equations below the statement of the problem in the derivation), we see that even for $f'(r)<0$ we would end up with the same equation as for $f'(r)>0$. Thus, problem 2 is apparently not a problem.
Problem 3: Is it really necessary to assume $ r \neq C_1$? I guess yes, as otherwise we would get $0=f'(r)^2(r^2-C_1^2) = C_1^2=r^2$ which is a contradiction to $r \neq 0.$ Am I right?
Problem 4: I don't like this $\pm $ there, as it could be that a solution has both positive and negative $f'(r)$, so the way I write the solution is not that general.
 A: A slightly different approach (in attachment) leads to :
$$f(r)=\frac{1}{c_1}\ln \left| c_1r \pm \sqrt{c_1^2r^2-1}\right|+c_2$$
which includes all real solutions, not forgeting  $f(r)=c_2$ at limit $c_1 \rightarrow \infty.$

A: Problem 1. Observe from the initial equation that a constant $f$ is a solution. Now if $f$ is not constant then $f'(r) \ne 0$ for some $r>0$  and hence $f'(s) \ne 0$ for all $s$ in some open interval $J$ containing $ r$.  So proceed to work within the largest possible such $J$  to find the the consequences of  a non-constant $f$. Problem 2 is not a problem. In the LHS of the equation just below your mention of Problem 2, change $\ln(r f'(r))$ to $\ln|r f'(r)|$and the equation is valid whether $f'(r)$ is positive or negative.Problem 3 is that you have deduced that dom$(f) \subset (C_1, \infty)$ for some $C_1 > 0$, otherwise the line above the mention of Problem 3 is false.So (referring to Problem 1) we know that $J \subset (C_1, \infty)$ for some $ C_1>0$.  . Problem 4 is not a problem either as you can see from the original equation, that if $f$ is a solution then so is $(-f)$. Finally the solutions you obtain imply that $J=(C_1, \infty)$  by the maximality of $J$. (Note that $f'(r) \ne 0$ for $r>C_1$ for these solutions.)  I hope this is not too muddled.   
